What is the axis y=7x^2+14x ?

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axis y=7x^2+14x

axis

y = 7 x^{2} + 14 x

x = - 1

Solution steps

y = 7 x^{2} + 14 x

Rewrite

y = 7 x^{2} + 14 x

in the standard form

4 \cdot \frac{1}{28} (y - (- 7)) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, - 7), p = \frac{1}{28}

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = - 1

f (x) = x^{2} + 6 x + 8

y = 2 x^{2} + 12 x + 10

y = 290 (8 x + 0.5 x^{2} - 1.4)

y = 6 x^{2} + 6 x - 36

y = 5 x^{2} - 4 x + 3